# "Recall the definition of the inclination of a line L:" #
# "If" \ \ m \ \ "is the slope of line" \ \ L \ \ "and" \ \ \theta \ \ "is the inclination of line" \ \ L, "then:" #
# \qquad \qquad \qquad \qquad \qquad \qquad tan(\theta) \ = \ m, \qquad \qquad "and" \qquad \quad \theta \in [ 0, \pi ). \qquad \qquad \qquad \quad \ \ (1) #
# "So, to find the inclination of a line, we can start by finding its" #
# "slope, and then use eqn. (1)." #
# "The line we are given is the line through the two points:" #
# \qquad \qquad \qquad \qquad \qquad ( - sqrt{3}, -1 ) \qquad \qquad "and" \qquad \qquad (0, -2 ). #
# "The slope of this line is:" #
# \qquad m \ = \ { -1 - (-2) } / { - sqrt{3} - 0 } \ = \ { -1 + 2 } / { - sqrt{3} } \ = \ { 1 } / { - sqrt{3} } \ = \ - { 1 } / { sqrt{3} }. #
# "The slope of this line is:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad m \ = \ - { 1 } / { sqrt{3} }. #
# "So, by eqn. (1), for the inclination" \ \ \theta \ \ "of this line:" #
# \qquad \qquad \qquad \qquad \qquad \quad \ tan(\theta) \ = \ - { 1 } / { sqrt{3} }, \qquad \qquad \theta \in [ 0, \pi ). #
# "As" \ \ tan(\theta) \ \ "is negative, and" \quad \theta \in [ 0, \pi ) , "we see that" #
# \ \theta \in"Quadrant II." #
# "So, we have:" #
# \qquad \qquad \qquad \qquad \qquad \quad \ tan(\theta) \ = \ - { 1 } / { sqrt{3} }, \qquad \qquad \theta \in"Quadrant II." \qquad \quad \quad \ \ (2) #
# "Recognizing the connection between the value above, and" #
# "the 30-60-90 right triangle, and remembering the 30-60-90" #
# "right triangle, we recall:" #
# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \ tan( 30^@ ) \ = \ { 1 } / { sqrt{3} }. #
# "So, a Quadrant II angle which has tangent value the opposite" #
# "of the above, is the Quadrant II angle that has reference angle" \ 30^@, "that is to say, the angle" \ 180^@ - 30^@: #
# \qquad \qquad \qquad \qquad \quad \quad \quad \ tan( 180^@ - 30^@ ) \ = \ - { 1 } / { sqrt{3} }. #
# \qquad \qquad :. \qquad \qquad \quad \quad \ \ tan( 150^@ ) \ = \ - { 1 } / { sqrt{3} }. #
# "So:" #
# \qquad \quad \ \ tan( 150^@ ) \ = \ - { 1 } / { sqrt{3} } \ = \ m \qquad "and" \qquad \quad 150^@ \in [ 0, \pi ). #
# "Thus:" #
# \qquad \qquad \qquad \theta \ = \ \150^@ \ \ "is the inclination of the given line." #
# "In radians:"\qquad \quad \150^@ \ = \ 150 \cdot \pi / 180 \ = \ 15/18 \pi \ = \ 5/6 \pi \ = \ { 5 \pi }/6. #
# "So, finally:" #
# \qquad \quad \ \theta \ = \ \150^@ \ = \ { 5 \pi }/6 \qquad "is the inclination of the given line." #
